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Thread: Probability Question

  1. #1 Probability Question 
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    Hi all, I've posted here along time ago under a different name, it was probably only 10 times or so. I forgot which email it was tied to so that account is gone.

    I'm in a pattern recognition class right now but my probability background is very weak.

    This is in relation to the infinite monkey problem. It is for homework but I'm not asking for a solution just a tip on where to look or even an explanation of a simpler version of this problem.

    The problem is actually on average how long would it take for a monkey to produce Hamlet if he had an editor who strikes away each incorrect stroke as soon as they happen. He types at 1 character per minute and there are 27 characters.

    I reduced this problem to this:
    Lets start with a fair 6 sided die. What is the AVERAGE number of rolls I would need to roll (any number) let's say a 5?

    Once I can answer that answering the original problem should be trivial.

    It seems like this is going to come down to some integral or infinite series. I haven't seen any reference to this in my notes yet and I'm trying to get a head start.


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    Forum Professor river_rat's Avatar
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    The expected number of throws is what you would expect it to be :

    The trick is to work out how to do that infinite sum


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    I guess it stands to reason that after a sufficient number of throws, 1/6 of all throws end up being a five. Therefore after n rolls you will have n/6 fives and 5n/6 rolls that are not fives, or an average of 5 failures between each success, therefore 6 rolls per success.
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    Quote Originally Posted by Harold14370 View Post
    I guess it stands to reason that after a sufficient number of throws, 1/6 of all throws end up being a five. Therefore after n rolls you will have n/6 fives and 5n/6 rolls that are not fives, or an average of 5 failures between each success, therefore 6 rolls per success.
    That makes sense Harold, I sort of thought that 6 would be closer to a worse case scenario.

    River Rat, where did that series come from, is it derived from somewhere else, is there a link that has some better explanation on the WHY of the math. Or any idea what area of probability this idea of average expectations comes from?
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    Forum Professor river_rat's Avatar
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    Quote Originally Posted by ThomasEE View Post
    River Rat, where did that series come from, is it derived from somewhere else, is there a link that has some better explanation on the WHY of the math. Or any idea what area of probability this idea of average expectations comes from?
    The probability of needing exactly n roles before your first success is where p is the probability of success i.e. you fail n-1 times, each failure having a probability of 1-p and you finally succeed with probability p.

    Knowing this it is easy to work out your expected number of throws.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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